physics =(
posted by Jordan on .
A person on a small, 7.00 meter high, hill aims a waterballoon slingshot straight at a friend hanging from a tree branch 4.00 meters above and 6.00 meters away. At the instant the water balloon is released at 7.00 m/s, the friend lets go and falls from the tree, hoping to avoid being hit by the water balloon. Calculate the following:
(a) the time for balloon to reach tree ______________s
(b) the height of balloon at that time ________________m
(c) the height of the falling friend at that time ___________________m
i don't even know where to begin. thanks!

Draw a picture.
It is not clear to me where the 7m hill comes in. If I interpreted it correctly, it is of no use except that it permits the water balloon and the person to fall freely below the ground!
See:
http://img689.imageshack.us/img689/5825/1260658746b.png
Let
D=horizontal distance between the two friends = 6m
V=vertical distance between the two friends = 4m
u= initial velocity of the balloon making an angle θ (theta) with the horizontal = 7 m/s
ux=horizontal component of initial velocity = ucos(θ)
uy=vertical component of initial velocity = usin(θ)
H1(t)=height above initial level of balloon at time t
= uy*t(1/2)gt²
H2(t)=height of friend measured from top of tree at time t
= 0 (1/2)gt²
1. Time to reach tree
T = D/ux
2. H1(T)=uy*T(1/2)gT²
3. H2(T)=(1/2)gT²
Note that H1 is measured from ground, and H2 is measured from top of tree, at 4 m above ground.
My calculations tell me that the balloon hits the poor friend. 
Forgot to mention:
θ=(theta) is the angle aimed directly at the friend, so
θ=arcTan(4m/6m)